When I was calculating Pi value with Viete's method, One method is single multiplication of $2^n\sqrt{2-\sqrt{2+\sqrt{2+... n\text{ times}}}}$. There is another way when we start with equilateral triangle it turns about to be $3\cdot 2^n\sqrt{2-\sqrt{2+\sqrt{2+\dots (n-1)\text{ times}}}}$ and at the end of the nest $\sqrt3$.
Comparing both this method separately returns smaller and smaller values for nested square roots. As everyone can understand even though both methods lead to infinitismal numbers both these small numbers are different and when multipled by 2^n or 3*2^n they result in $Pi$.
My doubt is both the infinitismal numbers are different and the numbers $2^n limit n \to\infty$ and $3*2^n limit n\to\infty$ are also different.
As we are deducing $\pi$ by 2 different methods, it must be true that, both infinitesimal numbers are different and also infinitely big numbers are also should be different. Can anyone clarify? So infinite number $2^n$ where $n$ approaches infinity and $3*2^n$ where $n$ approaches infinity should be different. Please clarify from this perspective